IJRET: International Journal of Research in Engineering and Technology
eISSN: 2319-1163 | pISSN: 2321-7308
BOUNDS ON DOUBLE DOMINATION IN SQUARES OF GRAPHS M. H. Muddebihal1, Srinivasa G2 1 2
Professor, Department of Mathematics, Gulbarga University, Karnataka, India, mhmuddebihal@yahoo.co.in Assistant Professor, Department of Mathematics, B. N. M. I. T , Karnataka, India, gsgraphtheory@yahoo.com
Abstract Let the square of a graph G , denoted by G 2 has same vertex set as in G and every two vertices u and v are joined in G 2 if and only if they are joined in G by a path of length one or two. A subset D of vertices of G 2 is a double dominating set if every vertex in G 2 is dominated by at least two vertices of D . The minimum cardinality double dominating set of G 2 is the double
( )
( ) were obtained in terms of elements of
domination number, and is denoted by γ d G 2 . In this paper, many bounds on γ d G 2 G . Also their relationship with other domination parameters were obtained.
Key words: Graph, Square graph, Double dominating set, Double domination number. Subject Classification Number: AMS-05C69, 05C70. --------------------------------------------------------------------***---------------------------------------------------------------------1. INTRODUCTION
The square of a graph G , denoted by G 2 has the same vertex set as in G and the two vertices u and v are joined
In this paper, we follow the notations of [1]. All the graphs considered here are simple, finite and connected. As usual p = V (G) and q = E (G ) denote the number of vertices
in G 2 if and only if they are joined in G by a path of length one or two (see [1], [2]).
and edges of G , respectively. In general, we use 〈 X 〉 to denote the subgraph induced by the set of vertices X and N (v ) and N [v ] denote the open
and closed neighborhoods of a vertex v , respectively. The notation α 0 ( G ) (α1 ( G ) ) is the minimum number of vertices(edges) is a vertex(edge) cover of G . Also β 0 ( G ) ( β1 ( G ) ) is the minimum number of vertices (edges) is a maximal independent set of vertex (edge) of G . Let deg ( v ) is the degree of a vertex v and as usual
δ ( G ) ( ∆ ( G ) ) denote the minimum (maximum) degree of
G . A vertex of degree one is called an end vertex and its neighbor is called a support vertex. Suppose a support vertex v is adjacent to at least two end vertices then it is called a strong support vertex. A vertex v is called cut vertex if removing it from G increases the number of components of G .
The distance between two vertices u and v is the length of the shortest u v - path in G . The maximum distance between any two vertices in G is called the diameter, denoted by diam ( G ) .
We begin by recalling some standard definitions from domination theory. A set S ⊆ V is said to be a double dominating set of G , if every vertex of G is dominated by at least two vertices of S . The double domination number of G is denoted by γ d (G ) and is the minimum cardinality of a double dominating set of G . This concept was introduced by F. Harary and T. W. Haynes [3]. A dominating set S ⊆ V (G ) is a restrained dominating set of G , if every vertex not in S is adjacent to a vertex in S and to a vertex in V − S . The restrained domination number of G , denoted by γ re ( G ) is the minimum cardinality of a restrained dominating set of G . This concept was introduced by G. S. Domke et. al.,[4]. A dominating set S ⊆ V (G ) is said to be connected dominating set of G , if the subgraph
S
is not
disconnected. The minimum cardinality of vertices in such a set is called the connected domination number of G and is denoted by γ c (G ) [5].
( ) is said to be a dominating set of G
A subset D ⊆ V G 2
2
,
if every vertex not in D is adjacent to some vertex in D .
( )
The domination number of G 2 , denoted by γ G 2 , is the
_______________________________________________________________________________________ Volume: 02 Issue: 09 | Sep-2013, Available @ http://www.ijret.org
454
IJRET: International Journal of Research in Engineering and Technology
eISSN: 2319-1163 | pISSN: 2321-7308
minimum cardinality of a dominating set of G 2 .
Theorem 2.2: For any connected graph G with p ≥ 3
Similarly, a dominating set D of G 2 is said to be total
p vertices, γ d G 2 ≤ + 1 . 2
( )
dominating set of G 2 , if for every vertex v ∈ V G 2 , there exists a vertex u ∈ D , u ≠ v , such that u is adjacent to v or if the subgraph D has no isolated vertex. The total domination number of G , denoted by γ t (G ) is the 2
2
minimum cardinality of total dominating set of G 2 . Domination parameters in squares of graphs were introduced by M. H. Muddebihal et. al.,(see [6] and [7]).
( )
Analogously, a subset D ⊆ V G 2
is said to be double
dominating set of G 2 , if every vertex in G 2 is dominated by at least two vertices of D . The double domination
( )
( )
( )
p Proof: For p ≤ 2 , γ d G 2 ≤/ . For p ≥ 3 , we prove 2 the result by induction process. Suppose p = V ≤ 3 in G ,
( )
p then γ d G 2 = . Assume that the result is true for any 2 graph with p -vertices. Let G be a graph with p + 1 vertices. Then by induction hypothesis, it follows that p + 1 γ d G2 ≤ . Hence the result is true for all graphs 2 with p ≥ 3 vertices by induction process.
( )
, is the minimum
Theorem 2.3: For any connected graph G with p ≥ 3
cardinality of a double dominating set of G 2 . In this paper,
vertices, γ d G 2 + γ G 2 ≤ p . Equality holds if and only if
number of G 2 , denoted by γ d G 2
( )
many bounds on γ d G 2
were obtained in terms of
elements of G . Also its relationship with other different domination parameters were expressed.
( ) ( )
G ≅ C3 , P3 .
Proof: Let S = {v1 , v2 ,..., vk } be the minimal set of vertices which covers all the vertices in G 2 . Clearly, S forms a dominating set of G 2 . Further, if there exists a vertex set
( )
V G 2 − S = V1 in G 2 . Then S ∪ V ' = D , where V ' ⊆ V1 in
2. RESULTS
( )
G 2 , be the set of vertices such that ∀ v ∈ V G 2 , there Theorem 2.1: a.
For any cycle C p , with p ≥ 3 vertices,
2, for p = 3. p 2 γ d C p = + 1, for p ≡ 0(mod 3) 3 p , otherwise. 3 b. For any complete graph K p , with p ≥ 2 vertices,
( )
( )
γ d K p2 = 2 . c.
For any star K1, n , with n ≥ 2 vertices,
(
)
γ d K1,2n = 2 . d.
For any wheel W p , with p ≥ 4 vertices,
( )=2.
e.
exists two vertices in S ∪ V = D . Further, since every '
vertex of G 2 are adjacent to at least two vertices of G 2 , clearly D forms a double dominating set of G 2 . Therefore,
( ) ( )
it follows that D ∪ S ≤ p . Hence γ d G 2 + γ G 2 ≤ p . Suppose, G ≅/ C3 , P3 . Then either 2 S ≠ D or D ∪ S < p , which gives a contradiction in both cases. Suppose, G ≅ C3 , P3 . Then in this case, D = 2 = 2 ⋅1 = 2 S .
( ) ( )
Clearly, D ∪ S = 3 = p . Therefore, γ d G 2 + γ G 2 = p . Theorem 2.4: For any connected
( )
( )
( p, q ) -
graph G ,
2γ G 2 ≤ γ d G 2 + 2 .
( )
Proof: Suppose S = {v1 , v2 ,..., vn } ⊆ V G 2 be the minimal set of vertices which covers all the vertices, such that dist ( u, v ) ≥ 3 for all {u, v} ∈ S . Then S forms a minimal
( )
dominating set of G 2 . Further, if for every v ∈ V G 2 ,
γ d W p2
there exists at least two vertices {u, w} ∈ S such that ∀ u , v ,
For any complete bipartite graph K p1 , p2 , with
N (v ) and N (u )
p1 + p2 = p vertices,
(
γ d K 2p , p 1
2
)=2.
( )
belongs to V G 2 − S . Then S itself is a
double dominating set of G 2 . Otherwise, there exists at least one vertex x ∈ N ( S ) such that S ∪ { x} = D forms a double
dominating
set
of
G2 .
Since
for
any
_______________________________________________________________________________________ Volume: 02 Issue: 09 | Sep-2013, Available @ http://www.ijret.org
455
IJRET: International Journal of Research in Engineering and Technology
( ) D +2 . Clearly, 2γ ( G ) ≤ γ ( G ) + 2 . S ≤ 2
graph G with p ≥ 2 , γ d G 2 ≥ 2 . Therefore, it follows that 2
2
2.5:
For
any
( ) + γ ( G ) ≤ p . Equality holds for
with p ≥ 3 vertices, γ d G
2
C3 , C4 , P3 , P4 , P5 , P7 .
Proof: Let F1 = {v1 , v2 ,..., vm } be the set of all non end
vertices in G . Suppose S = {v1 , v2 ,..., vk } ⊆ F1 , k ≤ m , be the minimal set of vertices which are at distance three covers all the vertices of G . Then S itself forms a minimal γ -set of G . Otherwise, there exists at least one vertex
v ∈ N [ S ] such that S ∪ {v} forms a minimal dominating set
of
G.
Now
G2 ,
in
since
v j ∈ D, 1 ≤ j ≤ i ,
there
exists
at
least
one
vertex
2
( p, q ) -graph G ,
connected
set D = {v1 , v2 ,..., vi } such that for every vertex
vk ∈ D, 1 ≤ k ≤ i . Also every vertex in G is adjacent to at
d
Theorem
vertex
eISSN: 2319-1163 | pISSN: 2321-7308
( )
V (G ) = V G 2 ,
let
I = {u1 , u 2 ,..., ui } be the set of all strong support vertices.
least two vertices of D in G 2 . Then D forms a minimal double dominating set of G 2 . Since
D ≥ 2 and the
diametral path includes at least two vertices. It follows that, diam ( G ) 2 D ≤ 2 p − diam ( G ) . Clearly, γ d G 2 ≤ p − . 2
( )
Theorem 2.7: For any nontrivial tree with p ≥ 3 vertices
( )
and m cut vertices, then γ d T 2 ≤ m + 1 , equality holds if T ≅ K1,n , n ≥ 2 .
Proof: Let B = {v1 , v2 ,..., vi } be the set of all cut vertices in
{
}
T with B = m . Suppose A = v1 , v2 ,..., v j , 1 ≤ j ≤ i be the
Suppose D = I ∪ F1' , where F1' ⊆ F1 − I be the minimum
set of cut vertices which are at a distance two from the end vertices of T and A ⊂ B . Now in T 2 , all the end vertices
set of vertices which covers all the vertices in G 2 , such that
are adjacent with ∀v j ∈ A and {B} − { A} . Now in T 2 ,
for every vertex v ∈ V G 2 , there exists at least two
since V (T ) = V T 2 , for every vertex v ∈ V T 2 , there
( )
vertices
{u, w} ∈ D where
∀ vi ∈ I
and
∀ v j ∈ F1' ∃{vk } ∈ V [G 2 ] − D has at least two neighbors which are either vi or v j . Then D forms a minimal double
G 2 . Therefore, it follows that
dominating set of
( )
D ∪ S ∪ {v} ≤ p . Hence γ d G 2 + γ ( G ) ≤ p . Suppose G ≅ C3 , C4 , P3 , P4 . Then in this case, D = 2 and
S = p − 2 . Clearly, it follows that D ∪ S = p . Therefore,
( )
γ d G 2 + γ (G ) = p . Suppose G ≅ P5 , P7 .
Then
in
this
case,
p D = 2
p and S = . Clearly, it follows that 2 p p D ∪ S = + . Therefore, γ d G 2 + γ ( G ) = p . 2 2
( )
( )
( )
exists at least two vertices {u , v} ∈ B = D in T 2 . Further, since D covers all the vertices in T 2 , D itself forms a minimal double dominating set of T 2 . Since every tree T contains at least one cut vertex, it follows that
( )
D ≤ m + 1 . Hence γ d T 2 ≤ m + 1 . Suppose T is isomorphic to a star K1,n . Then in this case,
D = 2 and
m =1.
( )
Therefore,
it
follows
that
( p, q) -graph
G,
γ d T 2 = m +1 . Theorem 2.8: For any connected
( )
( )
γ d G 2 ≤ γ t G 2 + ∆ (G ) . Proof: For p = 2 , the result follows immediately. Hence, let p ≥ 3 . Suppose V1 = {v1 , v2 ,..., vn } ⊆ V ( G ) be the set of all vertices with deg ( vi ) ≥ 2, 1 ≤ i ≤ n . Then there exists at
least one vertex v ∈ V1 of maximum degree ∆ ( G ) . Now in
( p, q ) diam ( G ) . ≤ p−
Theorem 2.6: For any connected
( )
p ≥ 3 vertices, γ d G 2
graph G with
2 diam ( G ) Proof: For p = 2 , γ d G 2 ≤/ p − . Hence 2 consider p ≥ 3 . Suppose there exists two vertices
( )
u , v ∈ V ( G ) , which constitutes the longest path in G . Then
( )
dist ( u , v ) = diam ( G ) . Since V ( G ) = V G 2 , there exists a
( )
G 2 , since V ( G ) = V G 2 , let D1 = {v1 , v2 ,..., vk } ⊆ V1 in G 2 . Suppose D1 covers all the vertices in G 2 and if the subgraph
D1
has no isolated vertex, then D1 itself is a
minimal total dominating set of G 2 . Otherwise, there exists
( )
a set D1 ∪ H , where H ⊆ V G 2 − D1 , forms a minimal total dominating set of G . Now let D = {v1 , v2 ,..., v j } ⊆ V1 2
in G 2 be the minimal set of vertices, which covers all the
( )
vertices in G 2 . Suppose ∀ v ∈ V G 2 , there exists at least
_______________________________________________________________________________________ Volume: 02 Issue: 09 | Sep-2013, Available @ http://www.ijret.org
456
IJRET: International Journal of Research in Engineering and Technology two vertices {u , w} ∈ D which are adjacent to at least one
( )
vertex of D and at least two vertices of V G 2 − D . Then D forms a γ d - set of G 2 . Otherwise D ∪ I , where
( )
I ⊆ V G 2 − D , forms a minimal double dominating set of G 2 . Clearly, it follows that
( )
( )
D ∪ I ≤ D1 ∪ H + ∆ ( G ) .
Theorem 2.9: For any connected
( )
exists at least two vertices
{u, w} ∈ D
in G 2 . Further, if
D covers all the vertices in G 2 , then D itself is a double dominating set of G . Clearly, it follows that D ≤ p − A + 1
( )
and hence γ d G 2 ≤ p − α 0 ( G ) + 1 . Suppose G ≅ K p . Then in this case, A = p − 1 and D = 2 . Clearly,
Therefore, γ d G 2 ≤ γ t G 2 + ∆ ( G ) .
eISSN: 2319-1163 | pISSN: 2321-7308
it
follows
D = p − A +1
that
( )
and
hence γ d G 2 = p − α 0 ( G ) + 1 .
( p, q) -graph
G,
γ ( G ) ≤ γ d G 2 . Equality holds if and only if γ ( G ) = 2 with diam ( G ) ≤ 3 .
Theorem 2.11: For any connected ( p, q) -graph G ,
( )
γ d G 2 ≤ γ t (G ) .
Proof: If V1 = {v1 , v2 ,..., vn } ⊆ V ( G ) be the set of vertices
Proof: Let K = {u1 , u2 ,..., un } ⊆ V ( G ) be the set of vertices
with deg ( vi ) ≥ 2, 1 ≤ i ≤ n . Then S = {v1 , v2 ,..., vk } ⊆ V1
such that N [ ui ] ∩ N u j = φ , where 1 ≤ i ≤ n , 1 ≤ j ≤ n .
forms a minimal dominating set of G . Now without loss of
Suppose there exists a minimal set K1 = {u1 , u2 ,..., u k } ∈ N ( K ) , such that the subgraph
generality
in
G2 ,
( )
V (G ) = V G 2 .
since
If
V2 = {v1 , v2 ,..., vk } be the set of vertices with deg ( vk ) < 2 . If V2 ∈ V (G ) , then the vertices which are at a distance at least two are adjacent to each vertex of V2 in G 2 . Hence
K ∪ K1 has no isolated vertex. Further, if K ∪ K1 covers all the vertices in G , then K ∪ K1 forms a minimal total
( )
dominating set of G . Since V ( G ) = V G 2 , there exists a
forms a minimal double
vertex set D = {v1 , v2 ,..., vm } ⊆ K ∪ K1 in G 2 , which covers
dominating set of G 2 . If V2 = φ , then S ∪ V3 = D where
all the vertices in G 2 and for every vertex v ∈ V G 2 , there
V3 ⊆ V1 forms a minimal double dominating set of G .
exists at least two vertices {u , w} ∈ D . Clearly, D forms a
Further, since every vertex in G 2 is adjacent to atleast two vertices of D , it follows that S ≤ D . Hence,
minimal double dominating set of G 2 . Therefore, it follows
S1 ∪ V2 = D
where S1 ⊆ S
2
( )
γ (G ) ≤ γ d G 2 . Suppose γ ( G ) ≠ 2 with diam ( G ) ≤ 3 . Then in this case
diam ( G ) = 1 and
hence,
( )
S =1.
S < D.
Clearly,
Further, if γ ( G ) = 2 with diam ( G ) ≤/ 3 . Then in this case,
( )
Clearly,
D > S .
Therefore,
( )
Hence γ ( G ) = γ d G 2 if and only if γ ( G ) = 2 with
Theorem 2.10: For any connected ( p, q) -graph G ,
( )
γ d G 2 ≤ p − α 0 ( G ) + 1 . Equality holds for K p . Proof:
Let
( p, q ) -graph
Theorem 2.12: For any connected
( )
G,
γ d G 2 ≤ β 0 ( G ) + 1 . Equality holds for K p . Suppose F = {u1 , u2 ,..., um } ⊆ V ( G ) be the set of all vertices
deg ( vi ) = 1, 1 ≤ i ≤ m .
with
A = {v1 , v2 ,..., vn } ⊆ V ( G ) ,
F ∪ F' ,
Then
where
F ⊆ V ( G ) − F , F ∉ N [ F ] forms a maximal independent '
γ ( G ) < γ d G 2 , again a contradiction. diam ( G ) ≤ 3 .
( )
that D ≤ K ∪ K1 . Hence γ d G 2 ≤ γ t ( G ) .
Proof: For p = 2 , the result is obvious. Hence let p ≥ 3 .
Therefore γ ( G ) < γ d G 2 , a contradiction.
diam ( G ) ≥ 4 .
( )
'
set of
vertices,
such that
( )
V ( G ) = V G 2 , let
D1 ∈ N ( F ) .
F ∪ F ' = β 0 ( G ) . Since
( )
D1 = {v1 , v2 ,..., vn } ⊆ V G 2 − F and
Suppose
( ) − D such
D2 ⊆ V G
2
1
that
D1 ∪ D2 = D forms a minimal set of vertices which covers all the vertices in G 2 . Further, if for every vertex
where
deg ( vi ) ≥ 2, 1 ≤ i ≤ n , be the minimum set of vertices which
( )
v ∈ V G 2 , there exists at least two vertices
{u, w} ∈ D .
covers all the edges of G , such that A = α 0 ( G ) . Now in
Then D forms a minimal double dominating set of G 2 . Since every graph G contains at least one independent
G 2 since V ( G ) = V G 2 , let D = {v1 , v2 ,..., vk } ⊆ A be the
vertex,
( )
( )
set of vertices such that for every vertex v ∈ V G 2 , there
it
follows
that
D ≤ F ∪ F ' +1.
Therefore,
( )
γ d G 2 ≤ β0 (G ) + 1 .
_______________________________________________________________________________________ Volume: 02 Issue: 09 | Sep-2013, Available @ http://www.ijret.org
457
IJRET: International Journal of Research in Engineering and Technology Suppose G ≅ K p . Then in this case, G contains exactly one independent vertex and by Theorem 2.1(b), it follows that
( )
γ d G 2 = β0 (G ) + 1 . Theorem
2.13:
( )
For
any
non-trivial
tree
T ,
γ d T 2 ≤ γ re (T ) + 1 . Proof: Let F = {v1 , v2 ,..., vn } ⊆ V (T ) be the set of vertices
with deg ( vi ) = 1, ∀ {vi } ∈ F , 1 ≤ i ≤ n . Suppose for every vertex v ∈ V (T ) − F , there exists a vertex u ∈ F and also a
x ∈ V (T ) − F . Then F itself is a restrained dominating set of T . Otherwise, there exists at least one vertex w ∈ V ( G ) − F , such that D ' = F ∪ {w} forms a minimal restrained dominating set of T . Let D = {u1 , u2 ,..., u k } ⊆ V − F in T 2 be the minimal set of
eISSN: 2319-1163 | pISSN: 2321-7308
[4]. G. S. Domke, J. H. Hattingh, S. T. Hedetniemi, R. C. Laskar and L. R. Markus, Restrained Domination in Graphs, Discrete Mathematics 203, 61–69, 1999. [5]. E. Sampathkumar and H. B. Walikar, The Connected Domination Number of Graphs, J.Math.Phy.Sci. 13, 607-613, 1979. [6]. M. H. Muddebihal, G. Srinivasa and A. R. Sedamkar, Domination in Squares of Graphs, Ultra Scientist, 23(3)A, 795–800, 2011. [7]
vertex
M. H. Muddebihal and G. Srinivasa, Bounds on Total Domination in Squares of Graphs, International Journal of Advanced Computer and Mathematical Sciences, 4(1), 67–74, 2013
( )
vertices which are chosen such that ∀ v ∈ V T 2 , there exists at least two vertices
{ y, z} ∈ D .
Further, since
2
D covers all the vertices in T , clearly D forms a minimal double dominating set of T 2 . Therefore, it follows that
D ≤ D ' + 1 due to the distance between vertices of T is
( )
one. Hence γ d T 2 ≤ γ re (T ) + 1 . Theorem
( )≤γ
γd G
2.14:
2
c
For
any
connected
graph
G,
(G ) +1 .
Proof: Suppose C = {v1 , v2 ,..., vn } ⊆ V ( G ) be the set of all cut vertices in G . Further, if C ∪ I , where I ∈ N ( C ) with
deg ( vi ) ≥ 2, ∀{vi } ∈ I be the minimal set of vertices which
covers all the vertices in G and if the sub graph C ∪ I
is
connected. Then C ∪ I forms a minimal connected dominating set of G . Let D = {v1 , v2 ,..., vk } be the minimal set of vertices which covers all the vertices in G 2 . Suppose
( )
for every vertex v ∈ V G 2 , there exists at least two vertices {u , w} ∈ D . Then D itself forms a minimal double dominating
set
of G 2 .
Therefore,
( )≤γ
D ≤ C ∪ I + 1 and hence γ d G
2
c
it
follows
that
(G ) +1 .
REFERENCES [1]. F. Harary, Graph Theory, Adison-Wesley, Reading, Mass., 1972. [2]. F. Harary and I. C. Ross, The square of a tree, Bell System Tech. J. 39, 641-647, 1960. [3]. F. Harary and T. W. Haynes, Double domination in graphs, Ars Combinatorica 55, 201-213, 2000.
_______________________________________________________________________________________ Volume: 02 Issue: 09 | Sep-2013, Available @ http://www.ijret.org
458